# Oscillatory behaviour of a higher-order dynamic equation

## Abstract

In this paper we are concerned with the oscillation of solutions of a certain more general higher-order nonlinear neutral-type functional dynamic equation with oscillating coefficients. We obtain some sufficient criteria for oscillatory behaviour of its solutions.

MSC:34N05.

## 1 Introduction

The calculus on time scales has been introduced in order to unify the theories of continuous and discrete processes and in order to extend those theories to a more general class of the so-called dynamic equations. In recent years there has been much research activity concerning the oscillation and non-oscillation of solutions of neutral dynamic equations on time scales.

In this paper we consider the higher-order nonlinear dynamic equation

${\left[y\left(t\right)+P\left(t\right)y\left(\tau \left(t\right)\right)\right]}^{{\mathrm{\Delta }}^{n}}+\sum _{i=1}^{m}{Q}_{i}\left(t\right){f}_{i}\left(y\left({\varphi }_{i}\left(t\right)\right)\right)=0,$
(1.1)

where $n\ge 2$, $P\left(t\right),{Q}_{i}\left(t\right)\in {C}_{rd}{\left[{t}_{0},\mathrm{\infty }\right)}_{\mathbb{T}}$ for $i=1,2,\dots ,m$; $P\left(t\right)$ is an oscillating function ($P\left(t\right):\mathbb{T}\to \mathbb{R}$), ${Q}_{i}\left(t\right)$ are positive real-valued functions for $i=1,2,\dots ,m$; ${\varphi }_{i}\left(t\right)\in {C}_{rd}{\left[{t}_{0},\mathrm{\infty }\right)}_{\mathbb{T}}$, ${\varphi }_{i}^{\mathrm{\Delta }}\left(t\right)>0$, the variable delays $\tau ,{\varphi }_{i}:{\left[{t}_{0},\mathrm{\infty }\right)}_{\mathbb{T}}\to \mathbb{T}$ with $\tau \left(t\right),{\varphi }_{i}\left(t\right) for all $t\in {\left[{t}_{0},\mathrm{\infty }\right)}_{\mathbb{T}}$, ${\varphi }_{i}\left(t\right)\to \mathrm{\infty }$ as $t\to \mathrm{\infty }$ for $i=1,2,\dots ,m$; $\tau \left(t\right)\to \mathrm{\infty }$ as $t\to \mathrm{\infty }$; ${f}_{i}\left(u\right)\in C\left(\mathbb{R},\mathbb{R}\right)$ are nondecreasing functions, $u{f}_{i}\left(u\right)>0$ for $u\ne 0$ and $i=1,2,\dots ,m$.

The purpose of the paper is to study oscillatory behaviour of solutions of equation (1.1). For the sake of convenience, the function $z\left(t\right)$ is defined by

$z\left(t\right)=y\left(t\right)+P\left(t\right)y\left(\tau \left(t\right)\right).$
(1.2)

## 2 Basic definitions and some auxiliary lemmas

A time scale $\mathbb{T}$ is an arbitrary nonempty closed subset of the real numbers . For $t\in \mathbb{T}$, we define the forward jump operator $\sigma :\mathbb{T}\to \mathbb{T}$ by

$\sigma \left(t\right):=inf\left\{s\in \mathbb{T}:s>t\right\}$

while the backward jump operator $\rho :\mathbb{T}\to \mathbb{T}$ is defined by

$\rho \left(t\right):=sup\left\{s\in \mathbb{T}:s

If $\sigma \left(t\right)>t$, we say that t is right-scattered, while if $\rho \left(t\right), we say that t is left-scattered. Also, if $\sigma \left(t\right)=t$, then t is called right-dense, and if $\rho \left(t\right)=t$, then t is called left-dense. The graininess function $\mu :\mathbb{T}\to \left[0,\mathrm{\infty }\right)$ is defined by

$\mu \left(t\right):=\sigma \left(t\right)-t.$

We introduce the set ${\mathbb{T}}^{\kappa }$ which is derived from the time scale $\mathbb{T}$ as follows. If $\mathbb{T}$ has left-scattered maximum m, then ${\mathbb{T}}^{\kappa }=\mathbb{T}-\left\{m\right\}$, otherwise ${\mathbb{T}}^{\kappa }=\mathbb{T}$.

Definition 1 

The function $f:\mathbb{T}\to \mathbb{R}$ is called rd-continuous provided it is continuous at right-dense points in $\mathbb{T}$ and its left-sided limits exist (finite) at left-dense points in $\mathbb{T}$.

Theorem 1 

Assume that $\nu :\mathbb{T}\to \mathbb{R}$ is strictly increasing and $\stackrel{˜}{\mathbb{T}}:=\nu \left(\mathbb{T}\right)$ is a time scale. Let $w:\stackrel{˜}{\mathbb{T}}\to \mathbb{R}$. If ${\nu }^{\mathrm{\Delta }}\left(t\right)$ and ${w}^{\stackrel{˜}{\mathrm{\Delta }}}\left(\nu \left(t\right)\right)$ exist for $t\in {\mathbb{T}}^{\kappa }$, then

${\left(w\circ \nu \right)}^{\mathrm{\Delta }}=\left({w}^{\stackrel{˜}{\mathrm{\Delta }}}\circ \nu \right){\nu }^{\mathrm{\Delta }},$

where we denote the derivative on $\stackrel{˜}{\mathbb{T}}$ by $\stackrel{˜}{\mathrm{\Delta }}$.

Definition 2 

Let $f:\mathbb{T}\to \mathbb{R}$ be a function. If there exists a function $F:\mathbb{T}\to \mathbb{R}$ such that ${F}^{\mathrm{\Delta }}\left(t\right)=f\left(t\right)$ for all $t\in {\mathbb{T}}^{\kappa }$, then F is said to be an antiderivative of f. We define the Cauchy integral by

Theorem 2 

Let u and v be continuous functions on $\left[a,b\right]$ that are Δ-differentiable on $\left[a,b\right)$. If ${u}^{\mathrm{\Delta }}$ and ${v}^{\mathrm{\Delta }}$ are integrable from a to b, then

${\int }_{a}^{b}{u}^{\mathrm{\Delta }}\left(t\right)v\left(t\right)\mathrm{\Delta }\left(t\right)+{\int }_{a}^{b}{u}^{\sigma }\left(t\right){v}^{\mathrm{\Delta }}\left(t\right)\mathrm{\Delta }\left(t\right)=u\left(b\right)v\left(b\right)-u\left(a\right)v\left(a\right).$

Let $\stackrel{˜}{\mathbb{T}}=\mathbb{T}\cup \left\{sup\mathbb{T}\right\}\cup \left\{inf\mathbb{T}\right\}$. If $\mathrm{\infty }\in \stackrel{˜}{\mathbb{T}}$, we call ∞ left-dense, and −∞ is called right-dense provided $-\mathrm{\infty }\in \stackrel{˜}{\mathbb{T}}$. For any left-dense ${t}_{0}\in \stackrel{˜}{\mathbb{T}}$ and any $\epsilon >0$, the set

${L}_{\epsilon }\left({t}_{0}\right)=\left\{t\in \mathbb{T}:0<{t}_{0}-t<\epsilon \right\}$

is nonempty, and so is ${L}_{\epsilon }\left(\mathrm{\infty }\right)=\left\{t\in \mathbb{T}:t>\frac{1}{\epsilon }\right\}$ if $\mathrm{\infty }\in \stackrel{˜}{\mathbb{T}}$.

Lemma 1 

Let $n\in \mathbb{N}$ and f be n-times differentiable on $\mathbb{T}$. Assume $\mathrm{\infty }\in \stackrel{˜}{\mathbb{T}}$. Suppose there exists $\epsilon >0$ such that

$f\left(t\right)>0,\phantom{\rule{1em}{0ex}}sgn\left({f}^{{\mathrm{\Delta }}^{n}}\left(t\right)\right)\equiv s\in \left\{-1,+1\right\}\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}\phantom{\rule{0.1em}{0ex}}t\in {L}_{\epsilon }\left(\mathrm{\infty }\right)$

and ${f}^{{\mathrm{\Delta }}^{n}}\left(t\right)\ne 0$ on ${L}_{\delta }\left(\mathrm{\infty }\right)$ for any $\delta >0$. Then there exists $p\in \left[0,n\right]\cap {\mathbb{N}}_{0}$ such that $n+p$ is even for $s=1$ and odd for $s=-1$ with

$\left\{\begin{array}{l}{\left(-1\right)}^{p+j}{f}^{{\mathrm{\Delta }}^{j}}\left(t\right)>0\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}\phantom{\rule{0.1em}{0ex}}t\in {L}_{\epsilon }\left(\mathrm{\infty }\right),j\in \left[p,n-1\right]\cap {\mathbb{N}}_{0},\\ {f}^{{\mathrm{\Delta }}^{j}}\left(t\right)>0\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}\phantom{\rule{0.1em}{0ex}}t\in {L}_{{\delta }_{j}}\left(\mathrm{\infty }\right)\phantom{\rule{0.25em}{0ex}}\left(\mathit{\text{with}}\phantom{\rule{0.1em}{0ex}}{\delta }_{j}\in \left(0,\epsilon \right)\right),j\in \left[1,p-1\right]\cap {\mathbb{N}}_{0}.\end{array}$

Lemma 2 

Let f be n-times differentiable on ${\mathbb{T}}^{{\kappa }^{n}}$, $t\in \mathbb{T}$, and $\alpha \in {\mathbb{T}}^{{\kappa }^{n}}$. Then with the functions ${h}_{k}$ defined as ${h}_{n}\left(t,s\right)={\left(-1\right)}^{n}{g}_{n}\left(s,t\right)$,

${h}_{0}\left(r,s\right)\equiv 1\phantom{\rule{1em}{0ex}}\mathit{\text{and}}\phantom{\rule{1em}{0ex}}{h}_{k+1}\left(r,s\right)={\int }_{s}^{r}{h}_{k}\left(\tau ,s\right)\mathrm{\Delta }s\phantom{\rule{1em}{0ex}}\mathit{\text{for}}\phantom{\rule{0.1em}{0ex}}k\in {\mathbb{N}}_{0},$

we have

$f\left(t\right)=\sum _{k=0}^{n-1}{h}_{k}\left(t,\alpha \right){f}^{{\mathrm{\Delta }}^{k}}\left(\alpha \right)+{\int }_{\alpha }^{{\rho }^{n-1}\left(t\right)}{h}_{n-1}\left(t,\sigma \left(\tau \right)\right){f}^{{\mathrm{\Delta }}^{n}}\left(\tau \right)\mathrm{\Delta }\tau .$

Lemma 3 

Let f be n-times differentiable on ${\mathbb{T}}^{{\kappa }^{n}}$ and $m\in \mathbb{N}$ with $m. Then we have, for all $\alpha \in {\mathbb{T}}^{{\kappa }^{n-1+m}}$ and $t\in {\mathbb{T}}^{{\kappa }^{m}}$,

${f}^{{\mathrm{\Delta }}^{m}}\left(t\right)=\sum _{k=0}^{n-m-1}{h}_{k}\left(t,\alpha \right){f}^{{\mathrm{\Delta }}^{k+m}}\left(\alpha \right)+{\int }_{\alpha }^{{\rho }^{n-m-1}\left(t\right)}{h}_{n-m-1}\left(t,\sigma \left(\tau \right)\right){f}^{{\mathrm{\Delta }}^{n}}\left(\tau \right)\mathrm{\Delta }\tau .$

Lemma 4 

Suppose f is n-times differentiable and ${g}_{k}$, $0\le k\le n-1$, are differentiable at $t\in {\mathbb{T}}^{{\kappa }^{n}}$ with

${g}_{k+1}^{\mathrm{\Delta }}\left(t\right)={g}_{k}\left(\sigma \left(t\right)\right)\phantom{\rule{1em}{0ex}}\mathit{\text{for all}}\phantom{\rule{0.1em}{0ex}}0\le k\le n-2.$

Then we have

${\left[\sum _{k=0}^{n-1}{\left(-1\right)}^{k}{f}^{{\mathrm{\Delta }}^{k}}{g}_{k}\right]}^{\mathrm{\Delta }}=f{g}_{0}^{\mathrm{\Delta }}+{\left(-1\right)}^{n-1}{f}^{{\mathrm{\Delta }}^{n}}{g}_{n-1}^{\sigma }.$

## 3 Main results

Lemma 5 Let f be n-times differentiable on ${\mathbb{T}}^{{\kappa }^{n}}$. If ${f}^{\mathrm{\Delta }}>0$, then for every λ, $0<\lambda <1$, we have

$f\left(t\right)\ge \lambda {\left(-1\right)}^{n-1}{g}_{n-1}\left(\sigma \left({T}^{\ast }\right),t\right){f}^{{\mathrm{\Delta }}^{n-1}}\left(t\right).$
(3.1)

Proof Let p, $0\le p\le n-1$, be the integer assigned to the function f as in Lemma 1. Because of ${f}^{\mathrm{\Delta }}>0$, we always have $p>0$. Furthermore, let ${T}^{\ast }\ge T$ be assigned to f by Lemma 1. Then, by using the Taylor formula on time scales, for every ${\rho }^{n-1}\left(t\right)\ge {T}^{\ast }$, we obtain

$f\left(t\right)\ge {\int }_{{T}^{\ast }}^{{\rho }^{n-1}\left(t\right)}{\left(-1\right)}^{n-1}{g}_{n-1}\left(\sigma \left(\tau \right),t\right){f}^{{\mathrm{\Delta }}^{n}}\left(\tau \right)\mathrm{\Delta }\tau .$
(3.2)

By using Theorem 2 and (3.2), we have

$f\left(t\right)\ge {\left(-1\right)}^{n-1}{g}_{n-1}\left(\sigma \left(t\right),t\right){f}^{{\mathrm{\Delta }}^{n-1}}\left(t\right)-{\int }_{{T}^{\ast }}^{{\rho }^{n-1}\left(t\right)}{\left(-1\right)}^{n-1}{g}_{n-1}\left(\sigma \left(\tau \right),t\right){f}^{{\mathrm{\Delta }}^{n-1}}\left(\tau \right)\mathrm{\Delta }\tau .$

Since f is n-times differentiable on ${\mathbb{T}}^{{\kappa }^{n}}$ and $m\in \mathbb{N}$ with $m, we have with n and f substituted by $n-m$ and ${f}^{{\mathrm{\Delta }}^{m}}$, respectively

${f}^{{\mathrm{\Delta }}^{m}}\left(t\right)\ge {\int }_{{T}^{\ast }}^{{\rho }^{n-m-1}\left(t\right)}{\left(-1\right)}^{n-m-1}{g}_{n-m-1}\left(\sigma \left(\tau \right),t\right){f}^{{\mathrm{\Delta }}^{n}}\left(\tau \right)\mathrm{\Delta }\tau .$

Also, for every ${\rho }^{n-1}\left(t\right)$, s with ${\rho }^{n-1}\left(t\right)\ge {T}^{\ast }$ and ${T}^{\ast }\le s\le t$, we have

${f}^{{\mathrm{\Delta }}^{m}}\left(s\right)\ge {\left(-1\right)}^{n-m-1}{g}_{n-m-1}\left(\sigma \left({T}^{\ast }\right),t\right){f}^{{\mathrm{\Delta }}^{n}}\left(t\right).$

This is obvious for $m=n-1$ and, when $m, it can be derived by applying the Taylor formula. Thus, for all $t\ge {T}^{\ast }$, we get

$f\left(t\right)\ge {\left(-1\right)}^{n-1}{g}_{n-1}\left(\sigma \left({T}^{\ast }\right),t\right){f}^{{\mathrm{\Delta }}^{n-1}}\left(t\right)$

and therefore the proof of the lemma can be immediately completed. □

The result of Lemma 5 is an extension of studies in  and . In order that the reader sees how the results in  (1.8.14) and  (Lemma 2) follow from (3.1), it is at this point only necessary to know that in the case $\mathbb{T}=\mathbb{Z}$, we have $\rho \left(t\right)=t-1$, $\sigma \left(t\right)=t+1$ and

${g}_{n-1}\left(\sigma \left({T}^{\ast }\right),t\right)=\frac{{\left(t-{T}^{\ast }-1\right)}^{\left(n-1\right)}}{\left(n-1\right)!},$

then we get the inequality in 

$u\left(t\right)\ge \frac{1}{\left(n-1\right)!}{\left(n-{n}_{1}\right)}^{\left(n-1\right)}{\mathrm{\Delta }}^{n-1}u\left({2}^{n-m-1}n\right);$

and in the case $\mathbb{T}=\mathbb{R}$, we have $\rho \left(t\right)=\sigma \left(t\right)=t$ and

${g}_{n-1}\left(\sigma \left({T}^{\ast }\right),t\right)=\frac{{\left(t-{T}^{\ast }\right)}^{\left(n-1\right)}}{\left(n-1\right)!},$

then we get the inequality in 

$u\left(t\right)\ge \frac{\vartheta }{\left(n-1\right)!}{\left(t\right)}^{n-1}{u}^{n-1}\left(t\right).$

For the cases $\mathbb{T}=\mathbb{Z}$ and $\mathbb{T}=\mathbb{R}$, some sufficient criterias for oscillatory behaviour of the solutions of the equation (1.1) were obtained by Bolat and Akın in  and , respectively. Furthermore, there might be other time scales that we cannot appreciate at this time due to our current lack of ‘real-world’ examples.

Theorem 3 Assume that n is odd and

(C1) ${lim}_{t\to \mathrm{\infty }}P\left(t\right)=0$,

(C2) ${\int }_{{t}_{0}}^{\mathrm{\infty }}{s}^{n-1}{\sum }_{i=1}^{m}{Q}_{i}\left(s\right)\mathrm{\Delta }s=\mathrm{\infty }$.

Then every bounded solution of equation (1.1) is either oscillatory or tends to zero as $t\to \mathrm{\infty }$.

Proof Assume that equation (1.1) has a bounded non-oscillatory solution $y\left(t\right)$. Without loss of generality, assume that $y\left(t\right)$ is eventually positive (the proof is similar when $y\left(t\right)$ is eventually negative). That is, $y\left(t\right)>0$, $y\left(\tau \left(t\right)\right)>0$ and $y\left({\varphi }_{i}\left(t\right)\right)>0$ for $t\ge {t}_{1}\ge {t}_{0}$ and $i=1,2,\dots ,m$. Assume further that $y\left(t\right)$ does not tend to zero as $t\to \mathrm{\infty }$. By (1.1), (1.2), we have for $t\ge {t}_{1}$

${z}^{{\mathrm{\Delta }}^{n}}\left(t\right)=-\sum _{i=1}^{m}{Q}_{i}\left(t\right){f}_{i}\left(y\left({\varphi }_{i}\left(t\right)\right)\right)<0.$
(3.3)

It follows that ${z}^{{\mathrm{\Delta }}^{j}}\left(t\right)$ ($j\in \left[0,n-1\right]\cap {\mathbb{N}}_{0}$) is strictly monotone and eventually of constant sign. Since $P\left(t\right)$ is an oscillatory function, there exists a ${t}_{2}\ge {t}_{1}$ such that if $t\ge {t}_{2}$, then $z\left(t\right)>0$. Since $y\left(t\right)$ is bounded, by virtue of (C1) and (1.2), there is a ${t}_{3}\ge {t}_{2}$ such that $z\left(t\right)$ is also bounded for $t\ge {t}_{3}$. Because n is odd and $z\left(t\right)$ is bounded, by Lemma 1, when $p=0$ (otherwise $z\left(t\right)$ is not bounded), there exists ${t}_{4}\ge {t}_{3}$ such that for $t\ge {t}_{4}$ we have ${\left(-1\right)}^{j}{z}^{{\mathrm{\Delta }}^{j}}\left(t\right)>0$, $j\in \left[0,n-1\right]\cap {\mathbb{N}}_{0}$.

In particular, since ${z}^{\mathrm{\Delta }}\left(t\right)<0$ for $t\ge {t}_{4}$, $z\left(t\right)$ is decreasing. Since $z\left(t\right)$ is bounded, we write ${lim}_{t\to \mathrm{\infty }}z\left(t\right)=L$ ($-\mathrm{\infty }). Assume that $0\le L<\mathrm{\infty }$. Let $L>0$. Then there exists a constant $c>0$ and a ${t}_{5}\ge {t}_{4}$ such that $z\left(t\right)>c>0$ for $t\ge {t}_{5}$. Since $y\left(t\right)$ is bounded, ${lim}_{t\to \mathrm{\infty }}P\left(t\right)y\left(\tau \left(t\right)\right)=0$ by (C1). Therefore, there exists a constant ${c}_{1}>0$ and a ${t}_{6}\ge {t}_{5}$ such that $y\left(t\right)=z\left(t\right)-P\left(t\right)y\left(\tau \left(t\right)\right)>{c}_{1}>0$ for $t\ge {t}_{6}$. So that we can find a ${t}_{7}$ with ${t}_{7}\ge {t}_{6}$ such that $y\left({\varphi }_{i}\left(t\right)\right)>{c}_{1}>0$ for $t\ge {t}_{7}$. From (3.3) we have

${z}^{{\mathrm{\Delta }}^{n}}\left(t\right)=-\sum _{i=1}^{m}{Q}_{i}\left(t\right){f}_{i}\left({c}_{1}\right)<0$
(3.4)

for $t\ge {t}_{7}$. If we multiply (3.4) by ${t}^{n-1}$ and integrate it from ${t}_{7}$ to t, we obtain

$F\left(t\right)-F\left({t}_{7}\right)\le -f\left({c}_{1}\right){\int }_{{t}_{7}}^{t}\sum _{i=1}^{m}{Q}_{i}\left(s\right){s}^{n-1}\mathrm{\Delta }s,$
(3.5)

where

$F\left(t\right)=\sum _{i=1}^{n-1}{\left(-1\right)}^{i+1}{\left({t}^{n-1}\right)}^{{\mathrm{\Delta }}^{i}}{z}^{{\mathrm{\Delta }}^{n-i}}\left({\sigma }^{i}\left(t\right)\right)$

and

${\sigma }^{i}\left(t\right)=\sigma \left({\sigma }^{i-1}\left(t\right)\right).$

Since ${\left(-1\right)}^{k}{z}^{{\mathrm{\Delta }}^{k}}\left(t\right)>0$ for $k=0,1,2,\dots ,n-1$ and $t\ge {t}_{4}$, we have $F\left(t\right)>0$ for $t\ge {t}_{7}$. From (3.5) we have

$-F\left({t}_{7}\right)\le -f\left({c}_{1}\right){\int }_{{t}_{7}}^{t}\sum _{i=1}^{m}{Q}_{i}\left(s\right){s}^{n-1}\mathrm{\Delta }s.$

By (C2) we obtain

$-F\left({t}_{7}\right)\le -f\left({c}_{1}\right){\int }_{{t}_{7}}^{t}\sum _{i=1}^{m}{Q}_{i}\left(s\right){s}^{n-1}\mathrm{\Delta }s=-\mathrm{\infty }$

as $t\to \mathrm{\infty }$. This is a contradiction. So, $L>0$ is impossible. Therefore, $L=0$ is the only possible case. That is, ${lim}_{t\to \mathrm{\infty }}z\left(t\right)=0$. Since $y\left(t\right)$ is bounded, by (C1) we obtain

$\underset{t\to \mathrm{\infty }}{lim}y\left(t\right)=\underset{t\to \mathrm{\infty }}{lim}z\left(t\right)-\underset{t\to \mathrm{\infty }}{lim}P\left(t\right)y\left(t\right)=0$

from (1.2).

Now let us consider the case of $y\left(t\right)<0$ for $t\ge {t}_{1}$. By (1.1) and (1.2),

${z}^{{\mathrm{\Delta }}^{n}}\left(t\right)=-\sum _{i=1}^{m}{Q}_{i}\left(t\right){f}_{i}\left(y\left({\varphi }_{i}\left(t\right)\right)\right)>0$

for $t\ge {t}_{1}$. That is, ${z}^{{\mathrm{\Delta }}^{n}}>0$. It follows that ${z}^{{\mathrm{\Delta }}^{j}}\left(t\right)$ ($j\in \left[0,n-1\right]\cap {\mathbb{N}}_{0}$) is strictly monotone and eventually of constant sign. Since $P\left(t\right)$ is an oscillatory function, there exists a ${t}_{2}\ge {t}_{1}$ such that if $t\ge {t}_{2}$, then $z\left(t\right)<0$. Since $y\left(t\right)$ is bounded, by (C1) and (1.2), there is a ${t}_{3}\ge {t}_{2}$ such that $z\left(t\right)$ is also bounded for $t\ge {t}_{3}$. Assume that $x\left(t\right)=-z\left(t\right)$. Then ${x}^{{\mathrm{\Delta }}^{n}}\left(t\right)=-{z}^{{\mathrm{\Delta }}^{n}}\left(t\right)$. Therefore, $x\left(t\right)>0$ and ${x}^{{\mathrm{\Delta }}^{n}}\left(t\right)<0$ for $t\ge {t}_{3}$. Hence, we observe that $x\left(t\right)$ is bounded. Since n is odd, by Lemma 1, there exists a ${t}_{4}\ge {t}_{3}$ and $p=0$ (otherwise $x\left(t\right)$ is not bounded) such that ${\left(-1\right)}^{j}{x}^{{\mathrm{\Delta }}^{j}}\left(t\right)>0$, $j\in \left[0,n-1\right]\cap {\mathbb{N}}_{0}$ and $t\ge {t}_{4}$. That is, ${\left(-1\right)}^{j}{z}^{{\mathrm{\Delta }}^{j}}\left(t\right)<0$, $j\in \left[0,n-1\right]\cap {\mathbb{N}}_{0}$ and $t\ge {t}_{4}$. In particular, for $t\ge {t}_{4}$ we have ${z}^{\mathrm{\Delta }}\left(t\right)>0$. Therefore, $z\left(t\right)$ is increasing. So, we can assume that ${lim}_{t\to \mathrm{\infty }}z\left(t\right)=L$ ($-\mathrm{\infty }). As in the proof of $y\left(t\right)>0$, we may prove that $L=0$. As for the rest, it is similar to the case of $y\left(t\right)>0$. That is, ${lim}_{t\to \mathrm{\infty }}y\left(t\right)=0$. This contradicts our assumption. Hence the proof is completed. □

Theorem 4 Assume that n is even and (C 1) holds. If the following condition is satisfied:

(C3) There is a function $\phi \left(t\right)$ such that $\phi \left(t\right)\in {C}_{rd}^{1}{\left[{t}_{0},\mathrm{\infty }\right)}_{\mathbb{T}}$. Moreover,

$\underset{t\to \mathrm{\infty }}{lim}sup{\int }_{{t}_{0}}^{t}\phi \left(s\right)\sum _{i=1}^{m}{Q}_{i}\left(s\right)\mathrm{\Delta }s=\mathrm{\infty }$

and

$\underset{t\to \mathrm{\infty }}{lim}sup{\int }_{{t}_{10}}^{t}\frac{{\left[{\phi }^{\mathrm{\Delta }}\left(s\right)\right]}^{2}}{\phi \left(s\right){g}_{n-2}^{\sigma }\left(\sigma \left({\varphi }_{i}\left(s\right)\right),{\varphi }_{i}\left(s\right)\right)}\mathrm{\Delta }s<\mathrm{\infty }$

for $\phi \left(t\right)$ and $i=1,2,\dots ,m$. Then every bounded solution of equation (1.1) is oscillatory.

Proof Assume that equation (1.1) has a bounded non-oscillatory solution $y\left(t\right)$. Without loss of generality, assume that $y\left(t\right)$ is eventually positive (the proof is similar when $y\left(t\right)$ is eventually negative). That is, $y\left(t\right)>0$, $y\left(\tau \left(t\right)\right)>0$ and $y\left({\varphi }_{i}\left(t\right)\right)>0$ for $t\ge {t}_{1}\ge {t}_{0}$. By (1.1) and (1.2), we have (3.3) for $t\ge {t}_{1}$. Then ${z}^{{\mathrm{\Delta }}^{n}}\left(t\right)<0$. It follows that ${z}^{{\mathrm{\Delta }}^{j}}\left(t\right)$ ($j\in \left[0,n-1\right]\cap {\mathbb{N}}_{0}$) is strictly monotone and eventually of constant sign. Since $P\left(t\right)$ is an oscillatory function, there exists a ${t}_{2}\ge {t}_{1}$ such that for $t\ge {t}_{2}$, we have $z\left(t\right)>0$. Since $y\left(t\right)$ is bounded, by (C1) and (1.2), there is a ${t}_{3}\ge {t}_{2}$, such that $z\left(t\right)$ is also bounded for $t\ge {t}_{3}$. Because n is even, by Lemma 1 when $p=1$ (otherwise $z\left(t\right)$ is not bounded), there exists ${t}_{4}\ge {t}_{3}$ such that for $t\ge {t}_{4}$ we have

${\left(-1\right)}^{j+1}{z}^{{\mathrm{\Delta }}^{j}}\left(t\right)>0,\phantom{\rule{1em}{0ex}}j\in \left[0,n-1\right]\cap {\mathbb{N}}_{0}.$
(3.6)

In particular, since ${z}^{\mathrm{\Delta }}\left(t\right)>0$ for $t\ge {t}_{4}$, $z\left(t\right)$ is increasing. Since $y\left(t\right)$ is bounded,

$\underset{t\to \mathrm{\infty }}{lim}P\left(t\right)y\left(\tau \left(t\right)\right)=0$

by (C1). Let $\delta >1$; i.e., there exists a ${t}_{5}\ge {t}_{4}$ such that by (1.2)

$y\left(t\right)=z\left(t\right)-P\left(t\right)y\left(\tau \left(t\right)\right)>\frac{1}{\delta }z\left(t\right)>0$

for $t\ge {t}_{5}$. We may find a ${t}_{6}\ge {t}_{5}$ such that for $t\ge {t}_{6}$ and $i=1,2,\dots ,m$,

$y\left({\varphi }_{i}\left(t\right)\right)>\frac{1}{\delta }z\left({\varphi }_{i}\left(t\right)\right)>0.$
(3.7)

From (3.3), (3.7) and the properties of f, we have

$\begin{array}{rl}{z}^{{\mathrm{\Delta }}^{n}}\left(t\right)& \le -\sum _{i=1}^{m}{Q}_{i}\left(t\right){f}_{i}\left(\frac{1}{\delta }z\left({\varphi }_{i}\left(t\right)\right)\right)\\ =-\sum _{i=1}^{m}{Q}_{i}\left(t\right)\frac{{f}_{i}\left(\frac{1}{\delta }z\left({\varphi }_{i}\left(t\right)\right)\right)}{z\left({\varphi }_{i}\left(t\right)\right)}z\left({\varphi }_{i}\left(t\right)\right)\end{array}$
(3.8)

for $t\ge {t}_{6}$. Since $z\left(t\right)>0$ is bounded and increasing, ${lim}_{t\to \mathrm{\infty }}z\left(t\right)=L$ ($0). By the continuity of f, we have

$\underset{t\to \mathrm{\infty }}{lim}\frac{{f}_{i}\left(\frac{1}{\delta }z\left({\varphi }_{i}\left(t\right)\right)\right)}{z\left({\varphi }_{i}\left(t\right)\right)}=\frac{{f}_{i}\left(\frac{L}{\delta }\right)}{L}>0.$

Then there is a ${t}_{7}\ge {t}_{6}$ such that for $t\ge {t}_{7}$, $i=1,2,\dots ,m$, we have

$\frac{{f}_{i}\left(\frac{1}{\delta }z\left({\varphi }_{i}\left(t\right)\right)\right)}{z\left({\varphi }_{i}\left(t\right)\right)}\ge \frac{{f}_{i}\left(\frac{L}{\delta }\right)}{2L}=\alpha >0.$
(3.9)

By (3.8), (3.9),

(3.10)

Set

$w\left(t\right)=\frac{{z}^{{\mathrm{\Delta }}^{n-1}}\left(t\right)}{z\left(\frac{1}{\delta }{\varphi }_{i}\left(t\right)\right)}.$
(3.11)

We know from (3.6) that there is a ${t}_{8}\ge {t}_{7}$ such that for a sufficiently large $t\ge {t}_{8}$, $w\left(t\right)>0$. Therefore, Δ-derivating (3.11) we obtain

$\begin{array}{rl}{w}^{\mathrm{\Delta }}\left(t\right)& =\frac{{z}^{{\mathrm{\Delta }}^{n}}\left(t\right)}{z\left({\delta }^{-1}{\varphi }_{i}\left(t\right)\right)}+{z}^{{\mathrm{\Delta }}^{n-1}}\left(\sigma \left(t\right)\right){\left(\frac{1}{z\left({\delta }^{-1}{\varphi }_{i}\left(t\right)\right)}\right)}^{\mathrm{\Delta }}\\ =\frac{{z}^{{\mathrm{\Delta }}^{n}}\left(t\right)}{z\left({\delta }^{-1}{\varphi }_{i}\left(t\right)\right)}-\frac{{\delta }^{-1}{\varphi }_{i}^{\mathrm{\Delta }}\left(t\right){z}^{{\mathrm{\Delta }}^{n-1}}\left(\sigma \left(t\right)\right){z}^{\mathrm{\Delta }}\left({\delta }^{-1}{\varphi }_{i}\left(t\right)\right)}{z\left({\delta }^{-1}{\varphi }_{i}\left(t\right)\right)z\left({\delta }^{-1}{\varphi }_{i}\left(\sigma \left(t\right)\right)\right)}\end{array}$
(3.12)
$\begin{array}{r}\le \frac{{z}^{{\mathrm{\Delta }}^{n}}\left(t\right)}{z\left({\delta }^{-1}{\varphi }_{i}\left(t\right)\right)}-\frac{{\delta }^{-1}{\varphi }_{i}^{\mathrm{\Delta }}\left(t\right){z}^{{\mathrm{\Delta }}^{n-1}}\left(\sigma \left(t\right)\right){z}^{\mathrm{\Delta }}\left({\delta }^{-1}{\varphi }_{i}\left(t\right)\right)}{{z}^{2}\left({\delta }^{-1}{\varphi }_{i}\left(\sigma \left(t\right)\right)\right)}\\ =\frac{{z}^{{\mathrm{\Delta }}^{n}}\left(t\right)}{z\left({\delta }^{-1}{\varphi }_{i}\left(t\right)\right)}-{\delta }^{-1}{w}^{\sigma }\left(t\right)\frac{{\varphi }_{i}^{\mathrm{\Delta }}\left(t\right){z}^{\mathrm{\Delta }}\left({\delta }^{-1}{\varphi }_{i}\left(t\right)\right)}{z\left({\delta }^{-1}{\varphi }_{i}\left(\sigma \left(t\right)\right)\right)}.\end{array}$
(3.13)

We know from (3.6) that there is a $t\ge {t}_{9}$ such that ${z}^{\mathrm{\Delta }}\left(t\right)>0$ and ${z}^{{\mathrm{\Delta }}^{n-1}}\left(t\right)>0$ for an even n. Since $z\left(t\right)>0$ is increasing $z\left({\delta }^{-1}{\varphi }_{i}\left(\sigma \left(t\right)\right)\right)\ge z\left({\delta }^{-1}{\varphi }_{i}\left(t\right)\right)$ for $i=1,2,\dots ,m$. Therefore, by Lemma 5,

$z\left({\delta }^{-1}{\varphi }_{i}\left(t\right)\right)\ge \lambda {\left(-1\right)}^{n-1}{g}_{n-1}\left(\sigma \left({\varphi }_{i}\left(t\right)\right),{\varphi }_{i}\left(t\right)\right){z}^{{\mathrm{\Delta }}^{n-1}}\left({\varphi }_{i}\left(t\right)\right).$
(3.14)

Then by Δ-derivating (3.14) and using ${g}_{n-1}^{\mathrm{\Delta }}\left(\sigma \left(t\right),t\right)={g}_{n-2}^{\sigma }\left(\sigma \left(t\right),t\right)$, we get

$\begin{array}{rl}{\left[z\left({\delta }^{-1}{\varphi }_{i}\left(t\right)\right)\right]}^{\mathrm{\Delta }}& \ge \lambda {\left(-1\right)}^{n-2}{g}_{n-1}^{\mathrm{\Delta }}\left(\sigma \left({\varphi }_{i}\left(t\right)\right),{\varphi }_{i}\left(t\right)\right){z}^{{\mathrm{\Delta }}^{n-1}}\left({\varphi }_{i}\left(t\right)\right)\\ \ge \lambda {\left(-1\right)}^{n-2}{g}_{n-2}^{\sigma }\left(\sigma \left({\varphi }_{i}\left(t\right)\right),{\varphi }_{i}\left(t\right)\right){z}^{{\mathrm{\Delta }}^{n-1}}\left({\varphi }_{i}\left(t\right)\right)\end{array}$

by Lemma 2

${z}^{\mathrm{\Delta }}\left({\delta }^{-1}{\varphi }_{i}\left(t\right)\right){\delta }^{-1}{\varphi }_{i}^{\mathrm{\Delta }}\left(t\right)\ge \lambda {\left(-1\right)}^{n-2}{g}_{n-2}^{\sigma }\left(\sigma \left({\varphi }_{i}\left(t\right)\right),{\varphi }_{i}\left(t\right)\right){z}^{{\mathrm{\Delta }}^{n-1}}\left({\varphi }_{i}\left(t\right)\right).$

Since ${\varphi }_{i}\left(t\right)\le t$, we obtain

${z}^{\mathrm{\Delta }}\left({\delta }^{-1}{\varphi }_{i}\left(t\right)\right)\ge \frac{\delta \lambda {\left(-1\right)}^{n-2}{g}_{n-2}^{\sigma }\left(\sigma \left({\varphi }_{i}\left(t\right)\right),{\varphi }_{i}\left(t\right)\right){z}^{{\mathrm{\Delta }}^{n-1}}\left(t\right)}{{\varphi }_{i}^{\mathrm{\Delta }}\left(t\right)}.$
(3.15)

Hence by (3.10), (3.14) and (3.15), we have

$\begin{array}{rl}{w}^{\mathrm{\Delta }}\left(t\right)\le & \frac{-\alpha {\sum }_{i=1}^{m}{Q}_{i}\left(t\right)z\left({\varphi }_{i}\left(t\right)\right)}{z\left({\delta }^{-1}{\varphi }_{i}\left(t\right)\right)}\\ -{\delta }^{-1}{w}^{\sigma }\left(t\right)\frac{\delta \lambda {\left(-1\right)}^{n-2}{g}_{n-2}^{\sigma }\left(\sigma \left({\varphi }_{i}\left(t\right)\right),{\varphi }_{i}\left(t\right)\right){z}^{{\mathrm{\Delta }}^{n-1}}\left(t\right)}{{\varphi }_{i}^{\mathrm{\Delta }}\left(t\right)}\frac{{\varphi }_{i}^{\mathrm{\Delta }}\left(t\right)}{z\left({\delta }^{-1}{\varphi }_{i}\left(\sigma \left(t\right)\right)\right)}\\ \le & \frac{-\alpha {\sum }_{i=1}^{m}{Q}_{i}\left(t\right)z\left({\varphi }_{i}\left(t\right)\right)}{z\left({\delta }^{-1}{\varphi }_{i}\left(t\right)\right)}\\ -{\delta }^{-1}{w}^{\sigma }\left(t\right)\frac{\delta \lambda {\left(-1\right)}^{n-2}{g}_{n-2}^{\sigma }\left(\sigma \left({\varphi }_{i}\left(t\right)\right),{\varphi }_{i}\left(t\right)\right){\varphi }_{i}^{\mathrm{\Delta }}\left(t\right)}{{\varphi }_{i}^{\mathrm{\Delta }}\left(t\right)}\frac{{z}^{{\mathrm{\Delta }}^{n-1}}\left(\sigma \left(t\right)\right)}{z\left({\delta }^{-1}{\varphi }_{i}\left(\sigma \left(t\right)\right)\right)}\\ \le & -\alpha \sum _{i=1}^{m}{Q}_{i}\left(t\right)-\lambda {\left(-1\right)}^{n-2}{g}_{n-2}^{\sigma }\left(\sigma \left({\varphi }_{i}\left(t\right)\right),{\varphi }_{i}\left(t\right)\right){\left({w}^{\sigma }\left(t\right)\right)}^{2},\end{array}$

and then

$\alpha \sum _{i=1}^{m}{Q}_{i}\left(t\right)\le -{w}^{\mathrm{\Delta }}\left(t\right)-\lambda {\left(-1\right)}^{n-2}{w}^{2}\left(t\right){g}_{n-2}^{\sigma }\left(\sigma \left({\varphi }_{i}\left(t\right)\right),{\varphi }_{i}\left(t\right)\right)$
(3.16)

for $t\ge {t}_{10}$. If we multiply (3.16) by $\phi \left(t\right)$ and integrate it from ${t}_{10}$ to t, we obtain by Theorem 2

$\begin{array}{rl}\alpha {\int }_{{t}_{10}}^{t}\phi \left(s\right)\sum _{i=1}^{m}{Q}_{i}\left(s\right)\mathrm{\Delta }s\le & -{\int }_{{t}_{10}}^{t}\phi \left(s\right){w}^{\mathrm{\Delta }}\left(s\right)\mathrm{\Delta }s\\ -{\int }_{{t}_{10}}^{t}\lambda {\left(-1\right)}^{n-2}\phi \left(s\right){w}^{2}\left(s\right){g}_{n-2}^{\sigma }\left(\sigma \left({\varphi }_{i}\left(s\right)\right),{\varphi }_{i}\left(s\right)\right)\mathrm{\Delta }s\\ \le & -\left[\phi \left(t\right)w\left(t\right)-\phi \left({t}_{10}\right)w\left({t}_{10}\right)-{\int }_{{t}_{10}}^{t}{\phi }^{\mathrm{\Delta }}\left(s\right){w}^{\sigma }\left(t\right)\mathrm{\Delta }s\right]\\ -{\int }_{{t}_{10}}^{t}\lambda {\left(-1\right)}^{n-2}\phi \left(s\right){w}^{2}\left(s\right){g}_{n-2}^{\sigma }\left(\sigma \left({\varphi }_{i}\left(s\right)\right),{\varphi }_{i}\left(s\right)\right)\mathrm{\Delta }s\\ \le & \phi \left({t}_{10}\right)w\left({t}_{10}\right)+{\int }_{{t}_{10}}^{t}{\phi }^{\mathrm{\Delta }}\left(s\right){w}^{\sigma }\left(t\right)\mathrm{\Delta }s\\ -\lambda {\int }_{{t}_{10}}^{t}\phi \left(s\right){w}^{2}\left(s\right){g}_{n-2}^{\sigma }\left(\sigma \left({\varphi }_{i}\left(s\right)\right),{\varphi }_{i}\left(s\right)\right)\mathrm{\Delta }s\\ \le & \phi \left({t}_{10}\right)w\left({t}_{10}\right)-\lambda {\int }_{{t}_{10}}^{t}\phi \left(s\right){g}_{n-2}^{\sigma }\left(\sigma \left({\varphi }_{i}\left(s\right)\right),{\varphi }_{i}\left(s\right)\right)\\ ×{\left[w\left(s\right)-\frac{{\phi }^{\mathrm{\Delta }}\left(s\right)}{2\lambda \phi \left(s\right){g}_{n-2}^{\sigma }\left(\sigma \left({\varphi }_{i}\left(s\right)\right),{\varphi }_{i}\left(s\right)\right)}\right]}^{2}\mathrm{\Delta }s\\ +{\int }_{{t}_{10}}^{t}\frac{{\left[{\phi }^{\mathrm{\Delta }}\left(s\right)\right]}^{2}}{4\lambda \phi \left(s\right){g}_{n-2}^{\sigma }\left(\sigma \left({\varphi }_{i}\left(s\right)\right),{\varphi }_{i}\left(s\right)\right)}\mathrm{\Delta }s\\ \le & \phi \left({t}_{10}\right)w\left({t}_{10}\right)+{\int }_{{t}_{10}}^{t}\frac{{\left[{\phi }^{\mathrm{\Delta }}\left(s\right)\right]}^{2}}{4\lambda \phi \left(s\right){g}_{n-2}^{\sigma }\left(\sigma \left({\varphi }_{i}\left(s\right)\right),{\varphi }_{i}\left(s\right)\right)}\mathrm{\Delta }s.\end{array}$

Therefore by (C3),

$\begin{array}{rl}\mathrm{\infty }& =\alpha \underset{t\to \mathrm{\infty }}{lim}sup{\int }_{{t}_{10}}^{t}\phi \left(s\right)\sum _{i=1}^{m}{Q}_{i}\left(s\right)\mathrm{\Delta }s\\ \le \phi \left({t}_{10}\right)w\left({t}_{10}\right)+\frac{1}{4\lambda }\underset{t\to \mathrm{\infty }}{lim}sup{\int }_{{t}_{10}}^{t}\frac{{\left[{\phi }^{\mathrm{\Delta }}\left(s\right)\right]}^{2}}{\phi \left(s\right){g}_{n-2}^{\sigma }\left(\sigma \left({\varphi }_{i}\left(s\right)\right),{\varphi }_{i}\left(s\right)\right)}\mathrm{\Delta }s\\ <\mathrm{\infty }.\end{array}$

Now let us consider the case of $y\left(t\right)<0$ for $t\ge {t}_{1}$. By (1.1) and (1.2), we have

${z}^{{\mathrm{\Delta }}^{n}}\left(t\right)=-\sum _{i=1}^{m}{Q}_{i}\left(t\right){f}_{i}\left(y\left({\varphi }_{i}\left(t\right)\right)\right)>0$

for $t\ge {t}_{1}$. That is, ${z}^{{\mathrm{\Delta }}^{n}}>0$. It follows that ${z}^{{\mathrm{\Delta }}^{j}}\left(t\right)$ ($j\in \left[0,n-1\right]\cap {\mathbb{N}}_{0}$) is strictly monotone and eventually of constant sign. Since $P\left(t\right)$ is an oscillatory function, there exists a ${t}_{2}\ge {t}_{1}$ such that $z\left(t\right)<0$ for $t\ge {t}_{2}$. Since $y\left(t\right)$ is bounded, by (C1) and (1.2), there is a ${t}_{3}\ge {t}_{2}$ such that $z\left(t\right)$ is also bounded for $t\ge {t}_{3}$. Assume that $x\left(t\right)=-z\left(t\right)$. Then ${x}^{{\mathrm{\Delta }}^{n}}\left(t\right)=-{z}^{{\mathrm{\Delta }}^{n}}\left(t\right)$. Therefore, $x\left(t\right)>0$ and ${x}^{{\mathrm{\Delta }}^{n}}\left(t\right)<0$ for $t\ge {t}_{3}$. Hence, we observe that $x\left(t\right)$ is bounded. Since n is odd, by Lemma 1, there exists a ${t}_{4}\ge {t}_{3}$ and $p=1$ (otherwise $x\left(t\right)$ is not bounded) such that ${\left(-1\right)}^{k}{x}^{{\mathrm{\Delta }}^{k}}\left(t\right)>0$, $k\in \left[0,n-1\right]\cap {\mathbb{N}}_{0}$ and $t\ge {t}_{4}$. That is, ${\left(-1\right)}^{k}{z}^{{\mathrm{\Delta }}^{k}}\left(t\right)<0$, $k\in \left[0,n-1\right]\cap {\mathbb{N}}_{0}$ and $t\ge {t}_{4}$. In particular, for $t\ge {t}_{4}$ we have ${z}^{\mathrm{\Delta }}\left(t\right)>0$. Therefore, $z\left(t\right)$ is increasing. For the rest of the proof, we can proceed the proof similarly to the case of $y\left(t\right)>0$. Hence, the proof is completed. □

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Correspondence to Deniz Uçar.

### Competing interests

The authors declare that they have no competing interests.

### Authors’ contributions

DU carried out the time scale studies, participated in the sequence alignment, drafted the manuscript, and have given final approval of the version to be published. YB carried out the preliminaries of the manuscript and participated in the sequence alignment. Each author have participated sufficiently in the work to take public responsibility for appropriate portions of the content. Authors read and approved the final manuscript.

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Uçar, D., Bolat, Y. Oscillatory behaviour of a higher-order dynamic equation. J Inequal Appl 2013, 52 (2013). https://doi.org/10.1186/1029-242X-2013-52 